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3 years ago

13

What is an angle that is complementary to DGE? (Please hell I’ll reward)

1 answer:

weqwewe [10]3 years ago

3 0

Answer:

<CGD

Step-by-step explanation:

Since m<CGB = right angle = 90°, therefore:

m<CGE = 90° (linear pair angles)

Complementary angles add up to 90°.

Thus, m<CGD + m<DGE = m<CGE = 90°.

Therefore, the angle that is complementary to <DGE is <CGD

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ILL MARK U BRAINLEST Find the circumference of the circle. Use 3.14 for 1. Round to the nearest hundredth, if necessary.

likoan [24]

Answer:

The circumference is 9 or approximately 28.26 inches.

Step-by-step explanation:

To calculate the circumference of a circle, multiply the diameter of the circle with π (pi). The circumference can also be calculated by multiplying 2×radius with pi (π=3.14).

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3 years ago

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A. 3<br> B. 1/3<br> C. -3<br> D. -1/3

guapka [62]

Answer:

D

Step-by-step explanation:

This is a negative slope since it is going from left to right so that leaves you with c or d.

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3 years ago

What is an equation of the line that passes through the points (0, 0)and (5, 2)?

beks73 [17]

Answer:

5y - 2x = 0

Step-by-step explanation:

Equation of line is given by,

y - y1 = ( x - x1 ) ( y2 - y1)/ (x2 - x1 )

y - 0 = ( x - 0 ) ( 2 - 0 )/ ( 5 - 0 )

y = 2x /5

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This is the required equation.

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3 years ago

Mackenzie spent a total of $17.50 on Saturday afternoon.She bought a movie ticket for $7.25 and snacks for $4.95.She spent the r

Helga [31]

17.50-7.25-4.95=5.30/2=2.65

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3 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

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3 years ago